Chemistry Homework Answer Key #3
Chapter 9 - Spontaneous Processes and Thermodynamic
Equilibrium
17.
1/3 CS2 (g) + O2 (g) ® 1/3 CO2 (g) + 2/3 SO2 (g) K = (K1)1/3
This is an example of a reaction for which the stoichiometric
coefficents have all been multiplied by a constant, n. In all such
cases the equilibrium for the "new" reaction is related to the stoichiometry
of the "old" reaction by, Knew = (Kold)n.
19. The reaction for which we are trying to find an equilibrium
constant is equal to the sum of the second reaction plus the reverse of
the first. Thus, the overall equilibrium constant would be is equal to
the product of the individual equilibrium constants, K2/K1.
|
|
|
|
________________________________________ |
|
|
|
23. For the reaction, PCl5 (g) « PCl3 (g) + Cl2 (g), the equilibrium expression is:
To calculate the equilibrium concentrations, a table of
partial pressures is set-up.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Substituting into the equilibrium expression yields,
Rearranging yields the quadratic, 0 = x2 + 2.15x + 0.192. Solving for x yields two values:
x = 0.6799, -2.829
Only the first value makes physical sense (the value of xin this problem can not be negative). The partial pressures of the gases are then,
PPCl5 = 0.215 atmPPCl3 = 0.680 atm
PCl2 = 0.680 atm
25. For the reaction, Br2 (g) +
I2 (g) « 2 BrI (g),
the equilibrium expression is:
To calculate the equilibrium concentrations, a table of
concentration values is set-up.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Substituting into the equilibrium expression yields,
Rearranging yields the quadratic, 0 = 318 x2 -28.98 x + 0.644. Solving for x yields two values:
x = 0.03842, 0.05271
Only the first value makes physical sense (the value of x can not exceed the initial partial pressure of I2, which is 0.04 atm). The partial pressures of the gases at equilibrium are then,
PBr2 = 0.0116 atmPI2 = 0.00160 atm
PIBr = 0.0768 atm
27. For the reaction, N2 (g) + O2
(g) « 2 NO (g), the equilibrium
expression is:
To calculate the equilibrium concentrations, a table of
concentration values is set-up.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Yuck. Given the extremely small value of K, it
is apparent that almost all of the NO will react to form N2
and O2. So, to make the math more tractable (as it stands x
will be very, very close to 0.11, so close that your calculator would not
be able to display the value to enough significant figures for you to see
that it is not 0.11). To make life a tiny bit easier, let's assume that
all of the NO does react, then x = 0.11 and the partial pressures
of N2 and O2 are 0.52 and 0.70. Now, let's perform
the equilibrium calculation using the following table.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Substituting into the equilibrium expression yields,
Rearranging yields the quadratic, 0 = (4 - 4.2 ´ 10-31) x2 + (5.124 ´ 10-31) x - 1.5288 ´ 10-31. Solving for x yields only one positive value:
The equilibrium partial pressures are then:
PNO = 3.90 ´ 10-16 atmPN2 = 0.52 atm
PO2 = 0.70 atm
33. For the reaction, 3 Al2Cl6
(g)
« 2 Al3Cl9
(g), the reaction quotient is:
The equilibrium constant as solved in problem 11 is 1.04 ´ 10-4. Therefore, given these partial pressures, Q > K; as the reaction approaches equilibrium it will move toward the left, that is, there will be a net consumption of Al3Cl9.
37. For the reaction, P4 (g) « 2 P2 (g), the reaction quotient is:
Since Q > K, the reaction will shift to right and
generate more reactant. To find the equilibrium partial pressures, we set
up the following table:
|
|
|
|
|
|
|
|
|
|
|
|
Rearranging yields the quadratic, 0 = 4 x2 - 8.612 x + 0.94. Solving for x yields two values:
x = 0.115, 2.038
Since x must be in the range, 0 < x < 1.00, we can exclude the second solution. The equilibrium partial pressures are then,
PP2 = 1.77 atmIf, after equilibrium is reached, the volume of the container is increased, the equilibrium will shift to the right, resulting in the generation of additional P2. To illustrate, let's assume the volume is doubled; the resulting pressures (by P1V1 = P2V2) before equilibrium is reestablished will then be: PP2 = 0.885 atm and PP4 = 2.56 atm. Applying the definition of Q gives,PP4 = 5.12 atm.
Since Q < K, more products must be generated
to reach equilibrium.
39. a) If additional product is added to the mixture, the equilibrium will shift to the left to partially consume the added product; more reactant will be generated.
b) If the volume is reduced, the reaction will shift to the right to partially offset the additional pressure (the right side has fewer gas molecules and a shift to the right will reduce the total pressure).47 & 51. Postponed until next week….c) If the reaction is cooled, the reaction will shift to the right to generate more heat to partially counteract the loss of "heat product". (Note that with exothermic reactions, you can view heat as one of the products; with endothermic reactions, you can view heat as one of the reactants).
d) If an inert substance is added to a mixture, there is no effect if the partial pressures of the other species present remain unchanged. In this case however, the total pressure is remaining constant, therefore the partial pressures of the other species are decreasing. This has the same effect as decreasing the partial pressures by increasing volume - the reaction will shift to the left to offset the decreased total partial pressure of the reacting species.
e) As stated above, the addition of an inert substance will have no effect if the partial pressures of the reacting species are unchanged. That is the case here.
60. Abbreviating the acid monomer as HA and the dimer as (HA)2, or HAHA, the reaction of interest is
2 HA « (HA)2
If the total pressure of the system at equilibrium is 0.725, we can assign PHA as x and P(HA)2 as 0.725 -x. Substituting into the equilibrium expression yields:
Only one value makes sense (x can not be negative
in this case). Therefore:
Thus the percentage of acetic acid that is dimerized is:
P(HA) = 0.327 atmP(HA)2 = 0.398 atm